If $y=A e^{m x}+B e^{n x}$,show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$.

Given,$y = A e^{mx} + B e^{nx}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = A \cdot m e^{mx} + B \cdot n e^{nx} = Am e^{mx} + Bn e^{nx}$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(Am e^{mx} + Bn e^{nx}) = Am^2 e^{mx} + Bn^2 e^{nx}$.
Now,substitute these into the expression $\frac{d^2y}{dx^2} - (m+n) \frac{dy}{dx} + mny$:
$= (Am^2 e^{mx} + Bn^2 e^{nx}) - (m+n)(Am e^{mx} + Bn e^{nx}) + mn(A e^{mx} + B e^{nx})$
$= Am^2 e^{mx} + Bn^2 e^{nx} - Am^2 e^{mx} - Bmn e^{nx} - Amn e^{mx} - Bn^2 e^{nx} + Amn e^{mx} + Bmn e^{nx}$
$= (Am^2 e^{mx} - Am^2 e^{mx}) + (Bn^2 e^{nx} - Bn^2 e^{nx}) + (-Bmn e^{nx} + Bmn e^{nx}) + (-Amn e^{mx} + Amn e^{mx})$
$= 0$.
Hence,proved.